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In general, if $R$ is any ring and $S$ is any right denominator set, the right globaldimension of the localization $R_S$ does not exceed that of $R$.

If $R$ is of finite global dimension and Noetherian, and $S$ is left and right denominator set, then $R_S$ and $R$ have the same dimension iff there is a simple $R$-module $M$ with $\operatorname{gldin}R=\operatorname{pdim}M$ and $M_S\neq0$. Therefore the dimension drops if this does not happen.

In the local commutative case, there is one simple module, so what you want from $v$ to decrease the dimension a lot is for it to turn one of the maps in the minimal resolution of the residue field $k$ of $R$ into a split one. For example, if in that resolution we have a piece looking like $$\cdots \to P_j\xrightarrow{\phi} P_{j-1}\to\cdots$$ there is a finite number of elements which, when inverted, turn $\phi$ into an isomorphism (this is called universal localization in some contexts) If $v$ is the product of these elements, you get a drop in the dimension.

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In general, if $R$ is any ring and $S$ is any right denominator set, the right globaldimension of the localization $R_S$ does not exceed that of $R$.

If $R$ is of finite global dimension and Noetherian, and $S$ is left and right denominator set, then $R_S$ and $R$ have the same dimension iff there is a simple $R$-module $M$ with $\operatorname{gldon}R=\operatorname{pdim}M$ \operatorname{gldin}R=\operatorname{pdim}M$and$M_S\neq0$. Therefoew Therefore the dimension drops if this does not happen. 1 In general, if$R$is any ring and$S$is any right denominator set, the right globaldimension of the localization$R_S$does not exceed that of$R$. If$R$is of finite global dimension and Noetherian, and$S$is left and right denominator set, then$R_S$and$R$have the same dimension iff there is a simple$R$-module$M$with$\operatorname{gldon}R=\operatorname{pdim}M$and$M_S\neq0\$. Therefoew the dimension drops if this does not happen.